Educational Codeforces Round 71 (Rated for Div. 2) A - There Are Two Types Of Burgers

Posted 2022-08-20 xwl3109377858

原文链接：https://www.cnblogs.com/xwl3109377858/p/11404050.html

Educational Codeforces Round 71 (Rated for Div. 2)

A - There Are Two Types Of Burgers

There are two types of burgers in your restaurant — hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet.

You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.

You have to answer t independent queries.

Input

The first line contains one integer t (1≤t≤100) – the number of queries.

The first line of each query contains three integers b, p and f (1≤b, p, f≤100) — the number of buns, beef patties and chicken cutlets in your restaurant.

The second line of each query contains two integers h and c (1≤h, c≤100) — the hamburger and chicken burger prices in your restaurant.

Output

For each query print one integer — the maximum profit you can achieve.

Example

input

3

15 2 3

5 10

7 5 2

10 12

1 100 100

100 100

output

40

34

0

Note

In first query you have to sell two hamburgers and three chicken burgers. Your income is 2⋅5+3⋅10=40.

In second query you have to ell one hamburgers and two chicken burgers. Your income is 1⋅10+2⋅12=34.

In third query you can not create any type of burgers because because you have only one bun. So your income is zero.

 

题意：题目意思是给你若干个面包，牛肉，鸡排。两个面包和一个牛肉可以做成一个牛肉面包，

两个面包和一个鸡排可以做成一个鸡肉面包，两种面包有两个价值，问你最大利益。

思路：因为两种面包原料两个面包一样，我们只需要看肉的不同，哪种面包的价值更大，

优先做哪种面包，原料有多可以再做另一种面包，就可以达到最大利益，具体看代码。

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<map>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 #include<stack>
11 #include<list>
12 using namespace std;
13 #define ll long long
14 const int mod=998244353;
15 const long long int inf=1e18+7;
16 //const int maxn=
17  
18 int main()
19 
20     ios::sync_with_stdio(false);
21     int T;
22     cin>>T;
23     int a,b,c;//a个面包，b个牛肉，c个鸡排 
24     int x,y;//牛肉面包价格x，鸡肉面包价格y 
25     while(T--) 
26     
27         cin>>a>>b>>c>>x>>y;
28         ll int sum=0;
29         if(x>=y)//做牛肉面包划算 
30         
31             sum+=min(a/2,b)*x;
32             a-=min(a/2,b)*2;
33             
34             if(a>0)//剩下的做鸡肉面包 
35                 sum+=min(a/2,c)*y;        
36         
37         else//做鸡肉面包划算 
38         
39             sum+=min(a/2,c)*y;
40             a-=min(a/2,c)*2;
41             
42             if(a>0)//剩下的做牛肉面包 
43                 sum+=min(a/2,b)*x;    
44         
45         cout<<sum<<endl;
46     
47     
48     return 0;
49